The answer is chloroplast
The Mass of oxygen in isolated sample is 8.6 g
<h3>What is the
Law of Constant composition?</h3>
The law of constant composition states that pure samples of the same compound contain the same element in the same ratio by mass irrespective of the source from which the compound is obtained.
Considering the given ascorbic acid samples:
Laboratory sample contains 1.50 gg of carbon and 2.00 gg of oxygen
mass ratio of oxygen to carbon is 2 : 1.5
Isolated sample will contain 2/1.5 * 6.45 g of oxygen.
Mass of oxygen in isolated sample = 8.6 g
In conclusion, the mass of oxygen is determined from the mass ratio of oxygen and carbon in the compound.
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Note that the complete question is given below:
A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains 1.50 g of carbon and 2.00 g of oxygen. Another sample of ascorbic acid isolated from citrus fruits contains 6.45 gg of carbon. According to the law of constant composition, how many grams of oxygen does this isolated sample contain?
Express the answer in grams to three significant figures.
8.47 g
Answer:
The equilibrium partial pressure of O2 is 0.545 atm
Explanation:
Step 1: Data given
Partial pressure of SO2 = 0.409 atm
Partial pressure of O2 = 0.601 atm
At equilibrium, the partial pressure of SO2 was 0.297 atm.
Step 2: The balanced equation
2SO2 + O2 ⇆ 2SO3
Step 3: The initial pressure
pSO2 = 0.409 atm
pO2 = 0.601 atm
pSO3 = 0 atm
Step 4: Calculate the pressure at the equilibrium
pSO2 = 0.409 - 2X atm
pO2 = 0.601 - X atm
pSO3 = 2X
pSO2 = 0.409 - 2X atm = 0.297
X = 0.056 atm
pO2 = 0.601 - 0.056 = 0.545 atm
pSO3 = 2*0.056 = 0.112 atm
Step 5: Calculate Kp
Kp = (pSO3)²/((pO2)*(pSO2)²)
Kp = (0.112²) / (0.545 * 0.297²)
Kp = 0.261
The equilibrium partial pressure of O2 is 0.545 atm
Answer:
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